Here is the limit I am trying to do
$$ \lim\limits_{x \to \infty} \frac{x^2 + \mathrm{e}^{4x}}{2x- \mathrm{e}^x} $$
Now, here first, I am trying to identify the indeterminate form so that I can use L'Hospital's rule. Numerator tends to $\infty$ as $x \to \infty $. But the denominator tends to $ \infty – \infty$ as $ x \to \infty$. So, indeterminate form would be
$$ \frac{\infty}{\infty – \infty} $$
So, how to approach this problem here ?
Best Answer
The denominator is indeed of the indeterminate form $\infty-\infty$, but simplifying it we get $$2x-e^x = e^x\left(\frac{2x}{e^x}-1\right)$$ and since $$\lim_{x\to\infty}\frac{2x}{e^x}=0$$ (e.g. by L'Hopital's), we see that the denominator tends to $-\infty$, so you can proceed with L'Hopital's for the original fraction as usual, as it is of the indeterminate form $\frac{\infty}{-\infty}$.